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kramer
3 years ago
15

What is the DMS measure of an angle that is 3.7 radians?

Mathematics
1 answer:
Nostrana [21]3 years ago
7 0
C. Because it equals to the degrees to the 3.7 radians
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Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part
sergey [27]

a) The limit of the position of particle Q when time approaches 2 is -\pi.

b) The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2.

c) The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}.

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for t = 2, we need to apply the following two <em>algebraic</em> substitutions:

u = \pi t (1)

k = 2\pi - u (2)

Then, the limit is written as follows:

x(t) =  \lim_{t \to 2} \frac{\sin \pi t}{2-t}

x(t) =  \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}

x(u) =  \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}

x(k) =  \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}

x(k) =  -\pi\cdot  \lim_{k \to 0} \frac{\sin k}{k}

x(k) = -\pi

The limit of the position of particle Q when time approaches 2 is -\pi. \blacksquare

b) The function velocity of particle Q is determined by the <em>derivative</em> formula for the division between two functions, that is:

v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}} (3)

Where:

  • f(t) - Function numerator.
  • g(t) - Function denominator.
  • f'(t) - First derivative of the function numerator.
  • g'(x) - First derivative of the function denominator.

If we know that f(t) = \sin \pi t, g(t) = 2 - t, f'(t) = \pi \cdot \cos \pi t and g'(x) = -1, then the function velocity of the particle is:

v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}

v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}

The velocity of particle Q is v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}} for all t \ne 2. \blacksquare

c) The vector <em>rate of change</em> of the distance between particle P and particle Q (\dot r_{Q/P} (t)) is equal to the <em>vectorial</em> difference between respective vectors <em>velocity</em>:

\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t) (4)

Where \vec v_{P}(t) is the vector <em>velocity</em> of particle P.

If we know that \vec v_{P}(t) = (0, 4), \vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right) and t = \frac{1}{2}, then the vector rate of change of the distance between the two particles:

\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)

\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)

\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)

The magnitude of the vector <em>rate of change</em> is determined by Pythagorean theorem:

|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}

|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}

The rate of change of the distance between particle P and particle Q at time t = \frac{1}{2} is \frac{4\sqrt{82}}{9}. \blacksquare

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

<em>Particle </em>P<em> moves along the y-axis so that its position at time </em>t<em> is given by </em>y(t) = 4\cdot t - 23<em> for all times </em>t<em>. A second particle, </em>Q<em>, moves along the x-axis so that its position at time </em>t<em> is given by </em>x(t) = \frac{\sin \pi t}{2-t}<em> for all times </em>t \ne 2<em>. </em>

<em />

<em>a)</em><em> As times approaches 2, what is the limit of the position of particle </em>Q?<em> Show the work that leads to your answer. </em>

<em />

<em>b) </em><em>Show that the velocity of particle </em>Q<em> is given by </em>v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}<em>.</em>

<em />

<em>c)</em><em> Find the rate of change of the distance between particle </em>P<em> and particle </em>Q<em> at time </em>t = \frac{1}{2}<em>. Show the work that leads to your answer.</em>

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

3 0
2 years ago
In the given figure find the value of x and y?<br>Show you works - Thank you!​
Vaselesa [24]

<u>EXPLANATION</u><u>:</u>

In ∆ ABC , ∠ABC = 40°

  • ∠CAB = x°
  • & ∠ACD = 50°

∠ACD is an exterior angle formed by extending BC to D

We know that

The exterior angle of a triangle formed by extending one side is equal to the sum of the opposite interior angles.

∠ACD = ∠CAB + ∠ABC

⇛50° = x° + 40°

⇛x° = 50°-40°

<h3>⇛x° = 10°</h3>

and

In ∆ ACD , AC = CD

⇛ ∠CDA = ∠CAD

Since the angles opposite to equal sides are equal.

Let ∠CDA = ∠CAD = A°

We know that

The sum of all angles in a triangle is 180°

In ∆ ACD,

∠CDA +∠CAD + ∠ACD = 180°

A°+A°+50° = 180°

⇛2A°+50° = 180°

⇛2A° = 180°-50°

⇛2A° = 130°

⇛A° = 130°/2

⇛A° = 65°

now,

∠CDA = ∠CAD = 65°

∠BAC + ∠CAD+y = 180°

Since angles in the same line

10°+65°+y = 180°

⇛75°+y =180°

⇛y = 180°-75°

<h3>⇛y = 105°</h3>

<u>Answer</u><u>:</u> Hence, the value of “x” & “y” will be 10° and 105° respectively.

4 0
2 years ago
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